I: X'->X the identity function with topology

[tomboi03]

Let X and X' denote a single set in the two topologies T and T', respectively. Let i:X'-> X be the identity function
a. Show that i is continuous <=> T' is finer than T.
b. Show that i is a homeomorphism <=> T'=T

This is all I've got.

According to the first statement... X $$\subset$$ T and X' $$\subset$$ T

a. if i is continuous... each open subset V of X' the set i^-1 is an open subset of X
T' is finer than T means... T $$\subset$$ T'.

i don't know where to go from here...

b. if i is a homeomorphism...
then... i is... a bijection therefore the function and the inverse function are continuous.
i and i^-1 are continuous. each open subset V of X' the set i^-1 is an open subset of X
if T' = T... then...

i don't know where to go from here either...

can someone help me out?

Thank You,
tomboi03

 

[lanedance]

don't know a heap about this stuff but would think you were pretty close, will have a try...

so
X $$\in$$ T
X' $$\in$$ T'

for a)
prove ->

i is the identity X' -> X
continuous means preimage of an open set is open

so given V$$\in$$ T (as the topology contains all the open sets)
then V' = i^-1.V is open
so V' is in T'
but
V' = i.V' = i. i^-1.V = V (little sketchy about this part??)
so
if
V$$\in$$ T
then
V is in T'

so T $$\subseteq$$ T'

now need to prove other direction...

 

[HallsofIvy]

T' "finer" than T means that any open set in T is an open set in T' (but there may be open sets in T' that are not in T).

Suppose T' is finer than T. If U is an open set in T, then i^-1 maps it back to itself, which is an open set in T' so the identity map is continuous.

Suppose i is continuous. It U is an open set in T, then i^-1 maps it back to itself, which, because i is continuous, is open and so in T'. Thus, every member of T is in T' and so T' is finer than T.